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МатематикаВопрос:
б) \(\frac{b+3}{b^3+9b}\)\(\cdot\) \(\frac{b+3}{b-3}+\frac{b-3}{b+3}\);
Ответ:
Преобразуем выражение:
- $$\frac{b+3}{b^3+9b} = \frac{b+3}{b(b^2+9)}$$
- $$\frac{b+3}{b-3}+\frac{b-3}{b+3} = \frac{(b+3)^2+(b-3)^2}{(b-3)(b+3)} = \frac{b^2+6b+9+b^2-6b+9}{(b-3)(b+3)} = \frac{2b^2+18}{(b-3)(b+3)} = \frac{2(b^2+9)}{(b-3)(b+3)}$$
Умножим первую дробь на вторую:
$$\frac{b+3}{b(b^2+9)} \cdot \frac{2(b^2+9)}{(b-3)(b+3)} = \frac{2}{b(b-3)}$$
Ответ: $$\frac{2}{b(b-3)}$$
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